If 0.98 g of an unknown was dissolved in 10.30 g of solvent and the resulting solution has a molality of 0.45 m, what is the molar mass of the unknown

Answers

Answer 1

Answer:

211.4g/mol.

Explanation:

Data obtained from the question includes:

Mass of unknown compound = 0.98g

Mass of solvent = 10.30g

Molality = 0.45 M

Next, we shall determine the number of mole of the unknown compound present in the solution.

This can be obtained as follow:

Molality = mole /kg of solvent

Mole of the unknown compound =.?

Mass of solvent = 10.30g = 10.30/1000 = 0.0103Kg

Molality = 0.45 M

Molality = mole /kg of solvent

0.45 = mole /0.0103

Cross multiply

Mole = 0.45 x 0.0103

Mole = 4.635×10¯³ mole

Therefore the mole of the unknown compound that dissolve in solution is 4.635×10¯³ mole

Now, we can obtain the molar mass of the unknown compound as follow:

Mole of the unknown compound = 4.635×10¯³ mole

Mass of unknown compound = 0.98g

Molar mass of the unknown compound =?

Mole = mass /Molar mass

4.635×10¯³ = 0.98 /Molar mass

Cross multiply

4.635×10¯³ x molar mass = 0.98

Divide both side by 4.635×10¯³

Molar mass = 0.98 / 4.635×10¯³

Molar mass = 211.4g/mol.

Therefore, the molar mass of the unknown compound is 211.4g/mol.

Answer 2

The molecular mass of the unknown has been 211.66 g/mol.

Molality can be defined as the moles of the solute per kg of solvent.

Molality can be expressed as:

Molality = [tex]\rm \dfrac{Mass\;of\;solute\;(g)}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{Mass\;of\;solvent\;(g)}[/tex] ......(i)

The given unknown has been the solute.

The mass of solute = 0.98 g.

The mass of solvent = 10.30 g.

The molality of the solution formed has been = 0.45 m.

Substituting the values in equation (i):

0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;\dfrac{1000}{10.30\;g}[/tex]

0.45 m = [tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}\;\times\;97.087[/tex]

[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex]  =  [tex]\rm \dfrac{0.45}{97.087}[/tex]

[tex]\rm \dfrac{0.98\;g}{molecular\;mass\;of\;solute}[/tex]  = 0.00463

Molecular mass of solute = [tex]\rm \dfrac{0.98}{0.00463}[/tex]

Molecular mass of solute = 211.66 g/mol.

The molecular mass of the unknown has been 211.66 g/mol.

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Related Questions

"Calculate the pH during the titration of 30.00 mL of 0.1000 M C3H7COOH(aq) with 0.1000 M LiOH(aq) after 29.9 mL of the base have been added. Ka of butanoic acid

Answers

Answer:

pH = 7.29

Explanation:

Ka of butanoic acid is 1.54x10⁻⁵

To obtain the pH of the solution you must use H-H equation for butanoic acid:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

Where pKa is defined as -log Ka = 4.81

Now, you need to find [C₃H₇COO⁻] and [C₃H₇COOH] concentrations (Also, you can find moles of each substance and replace them in the equation.

Butanoic acid reacts with LiOH, producing C₃H₇COO⁻, thus:

C₃H₇COOH + LiOH → C₃H₇COO⁻ + H₂O + Li⁺

Moles of both reactants, C₃H₇COOH and LiOH are:

C₃H₇COOH = 0.0300L ₓ (0.1000mol / L) = 0.003000moles of C₃H₇COOH

LiOH = 0.0299L ₓ (0.1000mol / L) = 0.00299 moles of LiOH.

That means moles of C₃H₇COO⁻ produced are 0.00299 moles.

And moles of C₃H₇COOH that remains in solution are:

0.00300 - 0.00299 = 0.00001 moles of C₃H₇COOH

Replacing in H-H equation:

pH = pKa + log₁₀ [C₃H₇COO⁻] / [C₃H₇COOH]

pH = 4.81 + log₁₀ [0.00299moles] / [0.00001moles]

pH = 7.29

The average bond length in C-C in benzene (C6H6) is between single and double bond lengths. This is due to which of the following effect?
1. Due to its refractive index2. Due to the possession of resonance3. Due to its H atoms4. Due to the H-bonds

Answers

Answer:

2. Due to the possession of resonance

Explanation:

In the benzene ring, the electrons that results in the bonds between the carbon atoms are delocalized. That is, they do not belong to a specific carbon atom. It is this unique feature that enables them to have a bond length between single and double bond lengths.

This feature is as a result of resonance.

The correct option is 2.

what is the electron configuration of the iodide ion?
A. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²
B. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶
C. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶
D. 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰

Answers

Answer:

the answer to this question is C

The electron configuration of the iodide ion is 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶. The correct option is option C.

What is electron configuration ?

The arrangement of an atom's or molecule's electrons in their respective atomic or molecular orbitals is known as its electron configuration; for instance, the electron configuration of the neon atom is 1s2 2s2 2p6.

According to electronic configurations, electrons move individually within each orbital while interacting with the average field produced by all other orbitals. The corrosion potential and reactivity of an atom may be calculated from its electron configuration. The electron configuration of the iodide ion is  1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶4d¹⁰5s²5p⁶.

Therefore, the correct option is option C.

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What is the absolute magnitude of the rate of change for [NH3] if the
rate of change for [Hz] is 9.00 M/s in the reaction 2 NH3(g) → N2(g) +
3 H2(g)?

Answers

Answer:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Explanation:

Hello,

In this case, we can write the law of mass action for the undergoing chemical reaction, based on the rates and the stoichiometric coefficients:

[tex]\frac{1}{-2}r_{NH_3} =\frac{1}{1} r_{N_2}=\frac{1}{3}r_{H_2}[/tex]

In such a way, knowing the rate of formation hydrogen (H₂), we can know the  rate of change of ammonia, that must be negative for consumption:

[tex]r_{NH_3} =\frac{-2}{3}r_{H_2}=\frac{-2}{3}*9.00\frac{M}{s} \\\\r_{NH_3} =-6.00\frac{M}{s}[/tex]

Nevertheless, the absolute magnitude will be positive:

[tex]r_{NH_3,abs} =6.00\frac{M}{s}[/tex]

Best regards.

A sample of gas is observed to effuse through a pourous barrier in 4.98 minutes. Under the same conditions, the same number of moles of an unknown gas requires 6.34 minutes to effuse through the same barrier. The molar mass of the unknown gas is:________.
g/mol.

Answers

Answer:

The molar mass of the unknown gas is [tex]\mathbf{ 51.865 \ g/mol}[/tex]

Explanation:

Let assume that  the gas is  O2 gas

O2 gas is to effuse through a porous barrier in time t₁ = 4.98 minutes.

Under the same conditions;

the same number of moles of an unknown gas requires  time t₂  =  6.34 minutes to effuse through the same barrier.

From Graham's Law of Diffusion;

Graham's Law of Diffusion states that, at a constant temperature and pressure; the rate of diffusion of a gas is inversely proportional to the square root of its density.

i.e

[tex]R \ \alpha \ \dfrac{1}{\sqrt{d}}[/tex]

[tex]R = \dfrac{k}{d}[/tex]  where K = constant

If we compare the rate o diffusion of two gases;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{d_2}{d_1}}[/tex]

Since the density of a gas d is proportional to its relative molecular mass M. Then;

[tex]\dfrac{R_1}{R_2}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

Rate is the reciprocal of time ; i.e

[tex]R = \dfrac{1}{t}[/tex]

Thus; replacing the value of R into the above previous equation;we have:

[tex]\dfrac{R_1}{R_2}={\dfrac{t_2}{t_1}}[/tex]

We can equally say:

[tex]{\dfrac{t_2}{t_1}}= {\sqrt{\dfrac{M_2}{M_1}}[/tex]

[tex]{\dfrac{6.34}{4.98}}= {\sqrt{\dfrac{M_2}{32}}[/tex]

[tex]M_2 = 32 \times ( \dfrac{6.34}{4.98})^2[/tex]

[tex]M_2 = 32 \times ( 1.273092369)^2[/tex]

[tex]M_2 = 32 \times 1.62076418[/tex]

[tex]\mathbf{M_2 = 51.865 \ g/mol}[/tex]

merits of modern periodic table?​

Answers

Answer:

Merits of modern periodic table:The wrong position of some elements like argon, potassium, cobalt and nickel due to atomic weights have been solved by arranging the elements in the order of increasing atomic number without changing their own places.The isotopes of some element have the same atomic numbers. Therefore, they find the same position in periodic table.It separates metals from non-metals.The groups of the table are divided into sub groups A and B due to their dissimilar properties which make the study of elements specific and easier.The representative and transition elements have been separated.

Hope this helps...

Good luck on your assignment...

Note the dynamic equilibrium in the opening photo which solution changes color when the pH of both solutions is increased explain?

Answers

Answer:

The colour of the orange solution becomes yellow.  

Explanation:

1. Before adding NaOH

Assume the picture showed a beaker of potassium chromate and one of potassium dichromate.

Both solutions are involved in the same equilibrium:

[tex]\rm\underbrace{\hbox{2CrO$_{4}^{2-}$(aq)}}_{\text{yellow}} +2H^{+}(aq) \rightleftharpoons \, \underbrace{\hbox{Cr$_{2}$O$_{7}^{2-}$}}_{\text{orange}} + H_{2}O[/tex]

The first beaker contains mostly chromate ions with a few dichromate ions.

The position of equilibrium lies to the left and the solution is yellow.

The second beaker contains mostly dichromate ions with a few chromate ions.

The position of equilibrium lies to the right and the solution is orange.

2. After adding NaOH

According to Le Châtelier's Principle, when we apply a stress to a system at equilibrium, the system will respond in a way that tends to relieve the stress.

Beaker 1

If you add OH⁻ to the equilibrium solution, it removes the H⁺ (by forming water).

The system responds by having the dichromate react with water to replace the H⁺.  

At the same time, the system forms more of the yellow chromate ion.

The position of equilibrium shifts to the left.

However, the solution is already yellow, so you see no change in colour.

Beaker 2

The reaction is the same as in Beaker 1.

This time, however, as the dichromate ion disappears, do does its orange colour.

Also, the yellow chromate is being formed and its yellow colour appears .

The colour changes from orange to yellow.

Consider the 1H NMR spectrum for the following compound:
CH3CH2CH3
Predict the first-order splitting patterns of the indicated proton. This question uses specific splitting patterns instead of the often used generic term "multiplet."
a. doublet of quartets
b. triplet of triplets
c. septet
d. quartet of quartets
e. quintet

Answers

Answer:

See explanation

Explanation:

In this case, we have to check the neighbors of each carbon in the molecule. In propane, we have two types of carbons (see figure 1) (blue and red ones). The red carbons are equivalent. (Both have the same neighbors). Now, we can analyze each carbon:

Blue carbon

In the blue carbon, we have 6 hydrogens neighbors (three on each methyl). If we follow the n+1 rule, we will have:

6+1= 7

For the blue carbon, we will have a Septet.

Red carbons

In the red carbon, we have only 2 neighbors (the carbon in the middle only have 2 hydrogens). If we follow the n+1 rule, we will have:

2+1=3

For the red carbon, we will have a triplet.

See figure 2

I hope it helps!

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of Bacteria and Archaea

Answers

Answer:

the answer is monerans

Explanation:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of Monera into the two kingdoms of Bacteria and Archaea.

What kingdom of Monera ?

Some biologists believed it made sense to classify prokaryotes as belonging to their own kingdom, the Monera. That served as the foundation for Richard Whittaker  and Lynn Margulis's five-kingdom proposal, which enhanced the Haeckel plan by include a kingdom of fungus.

Protists, protozoa, monera, fungi, and viruses have long been proposed as belonging to different kingdoms, but traditional evolutionists during the majority of the 20th century had given none of them any thought.

Later, the Monera kingdom was split into Eubacteria and Archaebacteria by Carl Woese .  Moreover, he divided the five kingdoms into three domains: Eukaryotes, Archaea, and Bacteria.

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Your question is incomplete. But your complete question is as follows:

When Carl Woese developed the modern system of classification, he broke the previous kingdom of into the two kingdoms of  _____  into  Bacteria and Archaea.

3. A student carries out the clay-catalyzed dehydration of cyclohexanol starting with 10 moles of cyclohexanol and obtains 500 mL of pure cyclohexene product. Calculate the student’s percent yield for this reaction. Show your work.

Answers

Answer:

[tex]49.45~%[/tex]

Explanation:

In this case, we have to start with the chemical reaction:

[tex]C_6H_1_2O~->~C_6H_1_0~+~H_2O[/tex]

So, if we start with 10 mol of cyclohexanol ([tex]C_6H_1_2O[/tex]) we will obtain 10 mol of cyclohexanol ([tex]C_6H_1_0[/tex]). So, we can calculate the grams of cyclohexanol if we calculate the molar mass:

[tex](6*12)+(10*1)=82~g/mol[/tex]

With this value we can calculate the grams:

[tex]10~mol~C_6H_1_0\frac{82~g~C_6H_1_0}{1~mol~C_6H_1_0}=820~g~C_6H_1_0[/tex]

Now, we have as a product 500 mL of [tex]C_6H_1_0[/tex]. If we use the density value (0.811 g/mL). We can calculate the grams of product:

[tex]500~mL\frac{0.811~g}{1~mL}=405.5~g[/tex]

Finally, with these values we can calculate the yield:

[tex]%~=~\frac{405.5}{820}x100~=~49.45%[/tex]%= (405.5/820)*100 = 49.45 %

See figure 1

I hope it helps!

Based on the data given, the percentage yield of the student's work is 49.45 %.

What is the equation of the reaction?

The equation of the clay-catalyzed dehydration of cyclohexanol is given below:

C₆H₁₂O ----> C₆H₁₀ + H₂O

From the equation of the reaction, 1 mole of cyclohexanol yields 1 mole of cyclohexene.

Therefore 10 moles of cyclohexanol should yield 10 moles of cyclohexene.

What is the moles of cyclohexene obtained?

First we determine the mass of cyclohexene obtained.

Mass = density * volume

volume of cyclohexene = 500 mL

density = 0.811 g/mL

mass of cyclohexene = 500 * 0.811

mass of cyclohexene = 405.5 g

Number of moles of cyclohexene = mass/molar mass

molar mass of cyclohexene = 82 g

moles of cyclohexene = 405.5 g/82

moles of cyclohexene = 4.945 moles

What is the percentage yield?

Percentage yield = actual yield /expected yield * 100%

Percentage yield = 4.945/10 * 100%

Percentage yield = 49.45%

Therefore, the percentage yield of the student's work is 49.45 %.

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The melting point of sodium chloride is 801°C. The melting point of chlorine is -101°C. Explain, in terms of structure and bonding, the difference between the melting points of these two substances.

Answers

Answer:

It's harder to melt sodium chloride because of its density and stability.

Explanation:

The bond between Na+ and Cl- is an ionic bond, meaning its an ionic compound. Ionic compounds have more inter molecular forces(forces involving multiple molecules). Chlorine has much less of these forces than NaCl and isn't as dense.

Answer:

It's harder to melt sodium chloride because of its density

Explanation:

The equilibrium constant, K, for the following reaction is 2.44×10-2 at 518 K: PCl5(g) PCl3(g) + Cl2(g) An equilibrium mixture of the three gases in a 15.3 L container at 518 K contains 0.300 M PCl5, 8.55×10-2 M PCl3 and 8.55×10-2 M Cl2. What will be the concentrations of the three gases once equilibrium has been reestablished, if the equilibrium mixture is compressed at constant temperature to a volume of 8.64 L?

Answers

Answer:

[PCl₅] = 0.5646M

[PCl₃] = 0.1174M

[Cl₂] = 0.1174M

Explanation:

In the reaction:

PCl₅(g) ⇄ PCl₃(g) + Cl₂(g)

K equilibrium is defined as:

K = 2.44x10⁻² = [PCl₃] [Cl₂] / [PCl₅]

The initial moles of each compound when volume is 15.3L are:

PCl₅ = 0.300mol/L×15.3L = 4.59mol

Cl₂ = 8.55x10⁻²mol/L×15.3L = 1.308mol

PCl₃ = 8.55x10⁻²mol/L×15.3L = 1.308mol

At 8.64L, the new concentrations are:

[PCl₅] = 4.59mol / 8.64L = 0.531M

[PCl₃] = 1.308mol / 8.64L = 0.151M

[Cl₂] = 1.308mol / 8.64L = 0.151M

At these conditions, reaction quotient, Q, is:

Q = [0.151M] [0.151M] / [0.531M]

Q = 4.29x10⁻²

As Q > K, the reaction will shift to the left producing more reactant, that means equilibrium concentrations are:

[PCl₅] = 0.531M + X

[PCl₃] = 0.151M - X

[Cl₂] = 0.151M - X

Where X is reaction coordinate.

Replacing in K expression:

2.44x10⁻² = [0.151M - X] [0.151M - X] / [0.531M + X]

1.296x10⁻² + 2.44x10⁻²X = 0.0228 - 0.302X + X²

0 = 9.84x10⁻³ - 0.3264X + X²

Solving for X:

X = 0.293 → False solution. Produce negative concentrations

X = 0.0336M → Right solution.

Replacing:

[PCl₅] = 0.531M + 0.0336

[PCl₃] = 0.151M - 0.0336

[Cl₂] = 0.151M - 0.0336

[PCl₅] = 0.5646M[PCl₃] = 0.1174M[Cl₂] = 0.1174M

If the charge on the chloride ion is -1, what is the charge on the magnesium ion in the compound MgCl2?

Answers

Answer:

+2

Explanation:

Because the charge of the chloride ion is negative, that means that the charge of the magnesium ion must be positive since cations and anions go together, not cation and cation nor anion and anion. Using the "reverse criss-cross method", since the subscript of Mg is 1, that means that this is the lowest whole number ratio so we don't need to worry about simplifying. Therefore, since the charge of Cl is 2, the answer is +2.

When 13.6 g of calcium chloride, CaCl2, was dissolved in 100.0 mL of water in a coffee cup calorimeter, the temperature rose from 25.00 °C to 31.75 °C. Determine the ∆Hrxn in kJ/mol for the reaction as written below. Assume that the density of the solutions is 1.000 g/mL, and that the specific heat capacity of the solutions is that of pure water. Must show your work (scratch paper) to receive credit.

Answers

Answer:

THE ENTHALPY OF SOLUTION IS 3153.43 J/MOL OR 3.15 KJ/MOL.

Explanation:

1. write out the variables given:

Mass of Calcium chloride = 13.6 g

Change in temperature = 31.75°C - 25.00°C = 6.75 °C

Density of the solution = 1.000 g/mL

Volume = 100.0 mL = 100.0 mL

Specific heat of water = 4.184 J/g °C

Mass of the water = unknown

2. calculate the mass of waterinvolved:

We must first calculate the mass of water in the bomb calorimeter

Mass = density  * volume

Mass = 1.000 * 100

Mass = 0.01 g

3. calculate the quantity of heat evolved:

Next is to calculate the quantity of heat evolved from the reaction

Heat = mass * specific heat of water * change in temperature

Heat = mass of water * specific heat *change in temperature

Heat = 13.6 g * 4.184 * 6.75

Heat = 13.6 g * 4.184 J/g °C * 6.75 °C

Heat = 384.09 J

Hence, 384.09J is the quantity of heat involved in the reaction of 13.6 g of calcium chloride in the calorimeter.

4. calculate the molar mass of CaCl2:

Next is to calculate the molar mas of CaCl2

Molar mass = ( 40 + 35.5 *2) = 111 g/mol

The number of moles of 13.6 g of CaCl2 is then:

Number of moles of CaCl2 = mass / molar mass

Number of moles = 13.6 g / 111 g/mol

Number of moles = 0.1225 mol

So 384.09 J of heat was involved in the reaction of 1.6 g of CaCl2 in a calorimter which translates to 0.1225 mol of CaCl2..

5. Calculate the enthalpy of solution in kJ/mol:

If 1 mole of CaCl2 is involved, the heat evolved is therefore:

Heat per mole = 384.09 J / 0.1225 mol

Heat = 3 135.43 J/mol

The enthalpy of solution is therefore 3153.43 J/mol or 3.15 kJ/mol.

The rate at which two methyl radicals couple to form ethane is significantly faster than the rate at which two tert-butyl radicals couple. Offer two explanations for this observation.

Answers

Answer:

1. stability factor

2. steric hindrance factor

Explanation:

stability of ethane is lesser to that of two tert-butyl, so ethane will be more reactive and faster.

ethane is less hindered and more reactive, while two tert-butyl is more hindered and less reactive

A 1.00 liter solution contains 0.31 M sodium acetate and 0.40 M acetic acid. If 0.100 moles of barium hydroxide are added to this system, indicate whether the following statements are TRUE or FALSE . (Assume that the volume does not change upon the addition of barium hydroxide.)

a. The number of moles of CH3COOH will remain the same.
b. The number of moles of CH3COO- will increase.
c. The equilibrium concentration of H3O+ will decrease.
d. The pH will decrease.
e. The ratio of [CH3COOH] / [CH3COO-] will remain the same.

Answers

Answer and Explanation:

The buffer solution is composed by sodium acetate (CH₃COONa) and acetic acid (CH₃COOH). Thus, CH₃COOH is the weak acid and CH₃COO⁻ is the conjugate base, derived from the salt CH₃COONa.

If we add a strong base, such as barium hydroxide, Ba(OH)₂, the base will dissociate completely to give OH⁻ ions, as follows:

Ba(OH)₂ ⇒ Ba²⁺ + 2 OH⁻

The OH⁻ ions will react with the acid (CH₃COOH) to form the conjugate base CH₃COO⁻.

Initial number of moles of CH₃COOH = 0.40 mol/L x 1 L = 0.40 mol

Initial number of moles of CHCOO⁻= 0.31 mol/L x 1 L = 0.31 mol

moles of OH- added: 2 OH-/mol x 0.100 mol/L x 1 L = 0.200 OH-

According to this, the following are the answers to the sentences:

a. The number of moles of CH₃COOH will remain the same ⇒ FALSE

The number of moles of CH₃COOH will decrease, because they will react with OH⁻ ions

b. The number of moles of CH₃COO⁻ will increase ⇒ TRUE

Moles of CH₃COO⁻ will be formed from the reaction of the acid (CH₃COOH) with the base (OH⁻ ions)

c. The equilibrium concentration of H₃O⁺ will decrease ⇒ FALSE

The equilibrium concentration of OH⁻ is increased

d. The pH will decrease⇒ FALSE

pKa for acetic acid is 4.75. We add the moles of base to the acid concentration and we remove the same number of moles from the conjugate base in the Henderson-Hasselbach equation to calculate pH:

[tex]pH= pKa + log \frac{[conjugate base + base]}{[acid - base]}[/tex]

pH = 4.75 + log (0.31 mol + 0.20 mol)/(0.40 mol - 0.20 mol) = 5.15

Thus, the pH will increase.

What volume of CH4(g), measured at 25oC and 745 Torr, must be burned in excess oxygen to release 1.00 x 106 kJ of heat to the surroundings

Answers

Answer:

[tex]V=27992L=28.00m^3[/tex]

Explanation:

Hello,

In this case, the combustion of methane is shown below:

[tex]CH_4+2O_2\rightarrow CO_2+2H_2O[/tex]

And has a heat of combustion of −890.8 kJ/mol, for which the burnt moles are:

[tex]n_{CH_4}=\frac{-1.00x10^6kJ}{-890.8kJ/mol}= 1122.6molCH_4[/tex]

Whereas is consider the total released heat to the surroundings (negative as it is exiting heat) and the aforementioned heat of combustion. Then, by using the ideal gas equation, we are able to compute the volume at 25 °C (298K) and 745 torr (0.98 atm) that must be measured:

[tex]PV=nRT\\\\V=\frac{nRT}{P}=\frac{1122.6mol*0.082\frac{atm*L}{mol*K}*298K}{0.98atm}\\\\V=27992L=28.00m^3[/tex]

Best regards.

HELPPP.
Which of the following is a property of matter?
O A. It takes up space.
OB. It is everywhere.
O C. It is constantly changing.
O D. It cannot be divided.

Answers

Answer:

a

Explanation:

Matter can be volume or density. So, this concludes that it is when it takes up space.

Answer: A.

Explanation:

it takes up space

What is the core charge of helium and why?

Answers

Answer:

Formula for effective nuclear charge is as follows. So, for He atom value of S = 0.30 because the electrons are present in 1s orbital. Therefore, calculate the effective nuclear charge for helium as follows. Thus, we can conclude that the effective nuclear charge for helium is 1.7

Explanation:

The effective nuclear charge experienced by a 1s electron in helium is +1.70.

Chemistry question. Image attached.

Answers

Answer:

The answer to your question is given below

Explanation:

The balanced equation for the reaction is given below:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

1. Writing an expression for the equilibrium constant, K.

The equilibrium constant, K for a reaction is simply the ratio of the concentration of the products raised to their coefficient to the concentration of the reactants raised to their coefficient.

Thus, we can write the equilibrium constant, K for the reaction as follow:

CaO(s) + CH4(g) + 2H2O(g) <=> CaCO3(s) + 4H2(g)

K = [CaCO3] [H2]⁴ / [CaO] [CH4] [H2O]²

2. Based on the value of K, more products will be in the equilibrium mixture since the value of K is a positive large number.

How many minutes would be required to electroplate 25.0 grams of chromium by passing a constant current of 4.80 amperes through a solution containing CrCl3

Answers

Answer:

483.27 minutes

Explanation:

using second faradays law of electrolysis

Two moles of copper (II) sulfate,CuSO4,contains blank moles of O

Answers

Answer:

8 mol

Explanation:

Step 1: Given data

Moles of copper (II) sulfate: 2 mol

Chemical formula of copper (II) sulfate: CuSO₄

Step 2: Establish the molar ratio of copper (II) sulfate to oxygen

According to the chemical formula, the molar ratio of copper (II) sulfate to oxygen is 1:4.

Step 3: Calculate the moles of O in 2 mol of CuSO₄

[tex]2molCuSO_4 \times \frac{4molO}{1molCuSO_4} = 8molO[/tex]

Which statement is true: Mg -> Mg2+ + 2e-

(1) Mg gains 2 electrons (2) Mg2+ loses 2 electrons (3) mg loses 1 electron (4) mg loses 2 electrons

Answers

Answer:

(2) Mg²⁺ loses 2 electrons.

Explanation:

Let's consider the following half-reaction.

Mg ⇒ Mg²⁺ + 2 e⁻

Magnesium is a metal that loses 2 electrons to fulfill the octet rule. When magnesium loses electrons, the oxidation number increases from 0 to +2. This kind of half-reaction is known as oxidation, and has to be accompanied by a reduction of another species.

List the following compounds in order from strongest acid to weakest acid. Rank the acids from strongest to weakest.
CH2CHCH2COOH CH2CH2CH2COOH CH3CHCH2COOH CH3CH2CH2COOH
Strongest Weakest

Answers

Answer:

CH3CH2CH2COOH<CH2(F)CH2CH2COOH<CH3CH(F)CH2COOH<CH2(F)CH(F)CH2COOH

Explanation:

We know that the presence of highly electronegative elements in carboxylic acid molecules lead to -I inductive effect. This implies that electrons are withdrawn along the chain towards the electronegative element. As electrons are withdrawn towards the electronegative element, the electron cloud of the carbonyl- hydrogen bond in the acid weakens and the hydrogen can now be easily lost as a proton, that is , the molecule becomes more acidic.

The -I inductive effect increases with increase in the number of electronegative elements present in the molecule and the proximity of the electronegative element to the carbonyl group. The closer the electronegative element is to the carbonyl group, the greater the acidity of the molecule since the -I inductive effect dies out with increasing distance from the carbonyl group. Also, the more the number of electronegative elements in the molecule, the greater the - I inductive effect and the greater the acidity of the molecule, hence the answer.

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Answers

Answer:

Option C. Will always.

Explanation:

A spontaneous reaction is a reaction that occurs without an external supply of heat.

This implies that spontaneous reaction will always occur as no external supply of heat is needed.

Argon gas has a boiling point of -197 °C. Which of the following diagrams best represents the
distribution of argon atoms in a steel sphere at -190 °C?

Answers

Answer:

(a)

Explanation:

Hello,

In this case, the temperature required to boil argon, it means, transform it from liquid to gas is -197 °C. In such a way, since the temperature inside the steel sphere is -190 °C, which is greater than the boiling point, we realize argon is gaseous, therefore, the molecules will be spread inside the sphere as they will be moving based on the kinetic theory of gases.

For that reason, answer is scheme (a).

Best regards.

You use 10.0 mL of solution A, 10.0 mL of solution B, and 70.0 mL of water for your first mixture. What is the initial concentration of KIO3

Answers

Complete Question

The complete question is shown on the first uploaded image

Answer:

The  initial concentration is  [tex]C_f = 0.0022 \ M[/tex]

Explanation:

From the question we are told that

    The volume of solution A is  [tex]V_i = 10.0 mL[/tex]

    The concentration of A is [tex]C_i = 0.0200 \ M[/tex]

    The volume of solution B  is  [tex]V_B = 10.0mL[/tex]

    The volume of water is  [tex]V_{w } = 70.0 mL[/tex]

Generally the law of dilution is mathematically represented as

             [tex]C_i * V_i = C_f * V_f[/tex]

Where  [tex]C_f[/tex] is the concentration of  the mixture

            [tex]V_f[/tex] is the volume of the mixture which is mathematically evaluated as

            [tex]V_f = 10 + 10 + 70[/tex]

           [tex]V_f = 90mL[/tex]

So  

      [tex]C_f = \frac{C_i * V_i}{ V_f}[/tex]

substituting values

       [tex]C_f = \frac{0.0200 * 10 }{90}[/tex]

       [tex]C_f = 0.0022 \ M[/tex]

Note the mixture obtained is  [tex]KIO_3[/tex]

Identify each of the following half-reactions as either an oxidation half-reaction or a reduction half-reaction.

H2(g) → 2H+(aq) + 2e- oxidation

Cu2+(aq) → + 2e-Cu(s) reduction

Write a balanced equation for the overall redox reaction. Use smallest possible integer coefficients.

Answers

Answer:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

Explanation:

In a redox reaction, one half-reaction is the oxidation (where the atom loss electrons) whereas the other reaction is the reduction (Where the atom is gaining electrons.

In the reactions:

H₂(g) → 2H⁺(aq) + 2e⁻ oxidation

Here, the reaction is written as the oxidation because the hydrogen H₂ is in oxidation state 0 and H⁺ in +1. That means each atom is loosing one electron.

Cu²⁺(aq) + 2e⁻ → Cu(s) reduction

And here, the Cu²⁺ is in +2 oxidation state and after the reaction is in Cu(s) 0 state. Thus, each atom is gaining 2 electrons.

The sum of both reactions is:

H₂(g) + Cu²⁺(aq) + 2e⁻ → 2H⁺(aq) + 2e⁻ + Cu(s)

Subtracting the electrons in both sides of the reaction:

H₂(g) + Cu²⁺(aq) → 2H⁺(aq) + Cu(s)

The element nitrogen forms a(n) _______ with the charge . The symbol for this ion is , and the name is . The number of electrons in this ion is .

Answers

Answer:

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10.

Explanation:

The element nitrogen is in the Group 15 in the Periodic Table, so it tends to gain 3 electrons (3 negative charges) to fill its valance shell with 8 electrons.

The element nitrogen forms an anion with the charge -3. The symbol for this ion is N³⁻, and the name is nitride. The number of electrons in this ion is 10 (the original 7 plus the 3 gained). It is isoelectronic with the gas Neon, which accounts for its stability.

A student followed the procedure of this experiment to determine the percent NaCl in a commercial bleaching solution that was found in the basement of an abandoned house. The student diluted 50.00 mL of the commercial bleaching solution to 250 mL in a volumetric flask, and titrated a 20-ml aliquot of the diluted bleaching solution. The titration required 35.46 mL of 0.1052M Na S,O, solution. A faded price label on the gallon bottle read $0.79. The density of the bleaching solution was 1.10 g ml.
(1) Calculate the mass of commercial bleaching solution titrated. 1.1 450=559
(2) Determine the percent NaOCl in the commercial bleaching solution.
(3) Calculate the mass of one gallon of the commercial bleaching solution.

Answers

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